The Third Annual Ubersite Geek Off (527 hits)
Category: HumorRating: 0.81 on 13 reviews (Rate this item) (V)
Submitted by Darth Famine <Dork Lord of the Uber> (View user info) at 2008-06-04 19:26:27 EDT
Thats right, the time has finally come.
For a year now I have reigned as geek supreme, but no more.
Take up your lightsabers and attempt to wrest the mantle of Dork lord of the Uber from me...
if
you
dare
So compose your list of geeky qualities and post them under the title "Geek-off Entry."
The Geek-off will go on until Friday the 13th. I will compile and judge the entries, rating them from 1-10 on the Geekometer. I will post the results on Saturday the 14th.
here are a few links for you to peruse
Last years geek off contest opener and winners announcment
http://www.ubersite.com/m/110330
http://www.ubersite.com/m/110579
User Reviews
Submitted by Darth_Famine (user info) at 2008-06-05 20:40:18 EDT (#)
Ranking: 0
heh, bring eet on :)
Submitted by messmind (user info) at 2008-06-05 12:19:42 EDT (#)
Ranking: 2
FOOOOOM!
Submitted by FALLEN (user info) at 2008-06-05 10:29:40 EDT (#)
Ranking: 2
fuck you you're going down.
R=2/Pi
http://www.geocities.com/frooha/grid/node2.html
Submitted by shadow (user info) at 2008-06-05 10:27:06 EDT (#)
Ranking: 1
meh, I'll play.
Submitted by centaur (user info) at 2008-06-05 04:10:19 EDT (#)
Ranking: -1
I KNOW NO-ONE OR NOTHING.
Submitted by Stagger_Lee (user info) at 2008-06-05 03:58:02 EDT (#)
Ranking: 0
I know darko, and he knows Hookhand alright.
Submitted by darko (user info) at 2008-06-04 23:02:46 EDT (#)
Ranking: 0
I know hookhand, and he is all about this
Submitted by Hookhand (user info) at 2008-06-04 22:48:39 EDT (#)
Ranking: 2
I am all about this
Submitted by simple_catalyst (user info) at 2008-06-04 20:28:17 EDT (#)
Ranking: 0
i spent my thursday night (and since I'm in a college town, that's equivalent to a friday for you 'real world folks') doing this, even though i did not have to.
\documentclass[11pt,epsf]{article}
\usepackage{graphics}
\usepackage{epsfig}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\theoremstyle{definition}
\newtheorem{theorem}{Theorem}
\newtheorem{example}{Example}
\newtheorem*{prob}{Problem}
\newcommand{\R}{\mathbb{R}}
\begin{document}
\pagestyle{empty}
\begin{center}
{\Large \textbf{Selected Solutions to Lay's \textit{Analysis With an Introduction to Proof}}}
\end{center}
\noindent
\begin{prob}[13.14]
Let $S$ be a bounded infinite set and let $x=\sup S$. Prove: If $x\notin S$, then $x\in S'$.
\noindent
\begin{proof}: Assume $x\notin S$. We know that for $\varepsilon > 0$, $x- \varepsilon$ is never an upper bound for $S$. Otherwise $(x- \varepsilon) \in \mathbb{R} \setminus S : x- \varepsilon > s, \forall s \in S$ and $x- \varepsilon \leq x$ which contradicts $x= \sup S$. Therefore, $N(x,\varepsilon) \cap S \neq \emptyset$ because $x- \varepsilon \in S$. Furthermore, since $x \notin S$, then $N^*(x,\varepsilon)\cap S \neq \emptyset$. Thus $x \in S'$.
\end{proof}
\end{prob}
\noindent
\begin{prob}[13.20]
Let $S$ and $T$ be subsets of $\mathbb{R}$. Prove the following:
\begin{description}
\item[] (a) $\mathrm{cl}(\mathrm{cl} (S))= \mathrm{cl}(S)$
By Theorem 13.17 (page 133) we know: (a) $\mathrm{cl}(S)$ is a closed set and that (b) $\mathrm{cl}(S)$ is closed iff $\mathrm{cl}(S)= \mathrm{cl}(\mathrm{cl}(S))$. Thus $\mathrm{cl}(\mathrm{cl} (S))= \mathrm{cl}(S).$
\item[] (b) $\mathrm{cl}(S \cup T)= (\mathrm{cl}(S)) \cup (\mathrm{cl}(T))$
\begin{proof}
Since $\mathrm{cl}(S \cup T) = (S \cup T) \cup (S \cup T)'$. And $x \in (S \cup T)'$ implies $N^*(x,\varepsilon) \cap (S \cup T) \neq \emptyset , \forall \varepsilon > 0$. Then by the distributiveness of sets, $N^*(x,\varepsilon) \cap (S \cup T) = (N^*(x,\varepsilon) \cap S) \cup (N^*(x,\varepsilon) \cap T)$. Thus $(N^*(x,\varepsilon) \cap S) \cup (N^*(x,\varepsilon) \cap T) \neq \emptyset$. Since $(N^*(x,\varepsilon) \cap S) \cup (N^*(x,\varepsilon) \cap T)$ can be re-written as $S' \cup T'$ we see that $x \in (S \cup T)' \Rightarrow x \in S' \cup T'$. Thus we have established that $(S \cup T)' \subseteq S' \cup T'$. For the reverse inclusion we assume $x \in S' \cup T'$. Then $(N^*(x,\varepsilon) \cap S) \neq \emptyset$ and $(N^*(x,\varepsilon) \cap T)\neq \emptyset$. Thus $(N^*(x,\varepsilon) \cap S) \cup (N^*(x,\varepsilon) \cap T) \neq \emptyset$. And we know that, $(N^*(x,\varepsilon) \cap S) \cup (N^*(x,\varepsilon) \cap T)= N^*(x,\varepsilon) \cap (S \cup T)$. Thus $N^*(x,\varepsilon) \cap (S \cup T)$ is non-empty, and $x \in (S \cup T)'$. This gives $(S \cup T)' \supseteq S' \cup T'$. Therefore $(S \cup T)' = S' \cup T'$. Using this equality gives that $\mathrm{cl}(S \cup T) = (S \cup T) \cup (S \cup T)' = (S \cup T) \cup (S' \cup T') = (S \cup S') \cup (T \cup T') = \mathrm{cl}(S) \cup \mathrm{cl}(T)$.
\end{proof}
\item[] (c) $\mathrm{cl}(S \cap T) \subseteq (\mathrm(S)) \cap(\mathrm{cl}(T))$
\begin{proof}
We know that $\mathrm{cl}(S \cap T)$ may be re-written as $(S \cap T) \cup (S \cap T)'$. Also $\mathrm{cl}(S) = S \cup S'$ and $\mathrm{cl}(T)= T \cup T'$. Thus we may write $(\mathrm{cl}(S)) \cap(\mathrm{cl}(T))$ as $(S \cup S') \cap (T \cup T')$. Again by distribution we have $(S \cup S') \cap (T \cup T')= S \cap (T \cup T') \cup S' \cap (T \cup T') = (S \cap T) \cup (S \cap T') \cup (S' \cap T) \cup (S' \cap T')$. It is clear that $(S \cap T) \cup (S \cap T)' \subseteq (S \cap T) \cup (S \cap T') \cup (S' \cap T) \cup (S' \cap T')$. Therefore we conclude that $x \in \mathrm{cl}(S \cap T) \Rightarrow x \in (\mathrm(S)) \cap(\mathrm{cl}(T))$.
\end{proof}
\item[] (d) Find an example to show that the equality in part (c) need not be strict.
Let $S=\{\frac{1}{n} : n \in \mathbb{N} \}$ and $T=(0,1)$. Then $S'=\{0\}$ and $T'=[0,1]$. Then $\mathrm{cl}(S \cap T) = \{\frac{1}{n} : n \in \mathbb{N} \setminus \{1\} \} \cup \{0\}$. But $\mathrm{cl}(S) \cap \mathrm{cl}(T) = \{\frac{1}{n} : n \in \mathbb{N} \} \cup \{0\}$. Clearly $\{1\} \in \mathrm{cl}(S) \cap \mathrm{cl}(T)$, but $\{1\} \notin \mathrm{cl}(S \cap T)$.
\end{description}
\end{prob}
\noindent
\begin{prob}[13.23]
For any set $S \subseteq \mathbb{R}$, let $S^o$ denote the union of all the open sets contained in S.
\begin{description}
\item[] (a)Prove that $S^o$ is an open set.
\begin{proof}
Let $S^o= \underset{A \in \mathcal{A}}{\bigcup}A$, and $ \mathcal{A} = \{A: A \subseteq S \wedge \mathrm{int}(A)=A\}$. If $x \in S^o$ then $\exists A \in \mathcal{A}$ such that, $x \in A$. Since $A$ is open $\exists N(x,\varepsilon): N(x,\varepsilon) \subseteq A$. Therefore, $N(x,\varepsilon) \subseteq A \subseteq S^o$ we conclude $S^o$ is open.
\end{proof}
\item[] (b)Prove that $S^o \subseteq S$ and for any $U: U \subseteq S \wedge \mathrm{int}(U)=U$ then $U \subseteq S^o$
\begin{proof}
Assume $x \in S^o$ then $\exists A \in \mathcal{A}$ such that $x \in A$. Since $\forall A$, $A \subseteq S$ we know $x \in S$. Thus $x \in S^o \Rightarrow x \in S$ and we conclude $S^o \subseteq S$. Furthermore, assume $x \in U$. Then by construction of $U$ we know $U \in \mathcal{A}$ and that if $x \in U$ then $x \in \underset{A \in \mathcal{A}}{\bigcup}A = S^o$ Thus $U \subseteq S^o$
\end{proof}
\item[] (c)Prove that $S^o = \mathrm{int}(A)$
\begin{proof}
If $x \in S^o$ then (by (a)) $N(x,\varepsilon) \subseteq S^o$ and (by (b)) $S^o \subseteq S$ such that $N(x,\varepsilon) \subseteq S$. Hence, $x \in \mathrm{int}(S)$ and $S^o \subseteq \mathrm{int}(S)$. Now, because $\mathrm{int}(S)$ is an open subset of $S$ we know $\mathrm{int}(S) \in \mathcal{A}$ and (like in (b) again) $\mathrm{int}(S) \subseteq S^o$. Thus we conclude $\mathrm{int}(S) = S^o$
\end{proof}
\end{description}
\end{prob}
\end{document}
Submitted by Shlongy (user info) at 2008-06-04 20:18:29 EDT (#)
Ranking: 0
This looks as fucking horrid as the last 14 Uber contests.
Submitted by matnotharry (user info) at 2008-06-04 19:58:03 EDT (#)
Ranking: 1
I'll see you on WoW
Submitted by Lib (user info) at 2008-06-04 19:42:29 EDT (#)
Ranking: 2
I love the cartoon! What a great idea.
Submitted by ChaosJester (user info) at 2008-06-04 19:31:53 EDT (#)
Ranking: 0
Lamest. Contest. Ever.